Question

If D, E and F are the mid-points of sides BC, CA and AB respectively of a $\Delta ABC$, then using coordinate geometry prove that Area of $\Delta DEF=\frac{1}{4}\left(Areaof\Delta ABC\right)$

Answer 1

Concept:
Application:

Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ be the vertices of $\Delta ABC$. Then, the coordinates of D, E and F are $\frac{x_2+x_3}{2},\left(\frac{y_2+y_3}{2}\right),\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})$ and $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ respectively.

${\Delta }_1=Areaof\Delta ABC=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

${\Delta }_2=Areaof\Delta DEF=\frac{1}{2}|\left(\frac{x_2+x_3}{2}\right)(\frac{y_1+y_3}{2}-\frac{y_1+y_2}{2})+\left(\frac{x_1+x_3}{2}\right)(\frac{y_1+y_2}{2}-\frac{y_2+y_3}{2})+\left(\frac{x_1+x_2}{2}\right)(\frac{y_2+y_3}{2}-\frac{y_1+y_3}{2})|$

$\Rightarrow {\Delta }_2=\frac{1}{8}|(x_2+x_3)(y_3-y_2)+(x_1+x_3)(y_1-y_3)+(x_1+x_2)(y_2-y_1)|$

$\Rightarrow {\Delta }_2=\frac{1}{8}|x_1\left(y_1\right)-(y_3+y_2-y_1)+x_2(y_3-y_2+y_2+y_1)+x_3(y_3-y_2+y_1-y_3)|$

$\Rightarrow {\Delta }_2=\frac{1}{8}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$\Rightarrow {\Delta }_2=\frac{1}{4}\left(Areaof\Delta ABC\right)=\frac{1}{4}{\Delta }_1$

Hence, Area of $\Delta DEF=\frac{1}{4}$ (Area of $\Delta ABC$)